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(D) The light beam enters the prism normally to face $AB$,so it passes undeviated into the prism.Let the angle of the prism at $A$ be $	heta$. The an

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$\sin 	heta > 8/9$

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(D) The light beam enters the prism normally to face $AB$,so it passes undeviated into the prism.Let the angle of the prism at $A$ be $	heta$. The angle of incidence at face $BC$ is $	heta$.For total internal reflection to occur at face $BC$,the angle of incidence $	heta$ must be greater than the critical angle $C$ for the glass-water interface.Thus,$\sin 	heta > \sin C$.The critical angle $C$ is given by $\sin C = \frac{\mu_{	ext{water}}}{\mu_{	ext{glass}}}$.Given $\mu_{	ext{glass}} = 1.5 = 3/2$ and $\mu_{	ext{water}} = 4/3$.So,$\sin C = \frac{4/3}{3/2} = \frac{4}{3} 	imes \frac{2}{3} = \frac{8}{9}$.Therefore,for total internal reflection,$\sin 	heta > 8/9$.

$A$ glass prism of refractive index $1.5$ is immersed in water (refractive index $4/3$) as shown in the figure. $A$ light beam incident normally on face $AB$ undergoes total internal reflection at face $BC$.

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